Homework Help:
Questions about Gravity and Acceleration of masses.

So as we all know the acceleration of an object is independent of it's mass (neglecting friction)...so if you dropped a rock and a car they should technically accelerate downward at the same rate. But what about if the object we were dropping was really massive, like so massive it would have its own gravitational field, wouldn't that field then affect the earth's gravitational field and pull the earth a bit towards it, and so this massive object wouldn't accelerate downward at the same rate as say a rock? Which would mean that acceleration due to gravity isn't actually completely independent of mass?
Where am I going wrong?

Right, I just used massive objects for a better example...but say you "dropped" Jupiter right above the earth, jupiter has a much stronger gravitational field, so wouldn't Jupiter remain stationary and the earth would accelerate towards it...so the acceleration of Jupiter would be zero as oppose to 9.8m/s^2? I realize this example is ridiculous but it's the best I could think of to get my point across.

Right, I just used massive objects for a better example...but say you "dropped" Jupiter right above the earth, jupiter has a much stronger gravitational field, so wouldn't Jupiter remain stationary and the earth would accelerate towards it...so the acceleration of Jupiter would be zero as oppose to 9.8m/s^2? I realize this example is ridiculous but it's the best I could think of to get my point across.

The force between them would be F=GMm/r2 where 'r' is the distance between the masses 'M' and 'm'. So essentially they'd both be under the same force by Newton's 3rd Law.

If you "dropped" Jupiter at some point relative to the Earth, both would feel the same force [tex]F=\frac{GMm}{r^2}[/tex], where M is the mass of Jupiter and m the mass of Earth. If you imagine some arbitrary point between Jupiter and the Earth, Jupiter would accelerate towards it with [tex]a=\frac{Gm}{r^2}[/tex] and the Earth would accelerate towards it with [tex]a=\frac{GM}{r^2}[/tex].